Wikipedia:Reference desk/Archives/Mathematics/2008 August 7

= August 7 =

Natural log integral
Does my working go wrong at any point?

$$\int_0^{\frac{\pi}{4}} \ln(1+ \tan \theta)\ d\theta$$

$$\theta = \pi/4 -\phi$$

$$\frac{d \theta}{d \phi} =-1$$

$$=-\int_{\frac{\pi}{4}}^0 \ln(1+ \tan (\frac{\pi}{4} -\phi))\ d\phi$$

$$=-\int_0^{\frac{\pi}{4}} \ln(1+ \tan (\frac{\pi}{4} -\phi))\ d\phi$$

$$=-\int_0^{\frac{\pi}{4}} \ln(1+ \frac{\tan{\frac{\pi}{4}}-\tan \phi}{1+{\tan{\frac{\pi}{4}}}{\tan \phi}})\ d\phi$$

$$=-\int_0^{\frac{\pi}{4}} \ln(1+ \frac{1-\tan \phi}{1+{\tan \phi}})\ d\phi$$

$$=-\int_0^{\frac{\pi}{4}} \ln(\frac{1+\tan \phi}{1+ \tan \phi}+ \frac{1-\tan \phi}{1+{\tan \phi}})\ d\phi$$

$$=-\int_0^{\frac{\pi}{4}} \ln(\frac{1+\tan \phi +1 - \tan \phi}{1+ \tan \phi})\ d\phi$$

$$=-\int_0^{\frac{\pi}{4}} \ln(\frac{2}{1+ \tan \phi})\ d\phi$$

$$=-\int_0^{\frac{\pi}{4}} \ln2 - \ln(1+ \tan \phi)\ d\phi$$

$$= \int_0^{\frac{\pi}{4}} \ln(1+ \tan \phi) - \ln2\ d\phi$$

$$ \Rightarrow \int_0^{\frac{\pi}{4}} \ln(1+ \tan \theta)\ d\theta = \int_0^{\frac{\pi}{4}} \ln(1+ \tan \phi) - \ln2\ d\phi$$

The whole point of the question is that I have to show that

$$\int_0^{\frac{\pi}{4}} \ln(1+ \tan \theta)\ d\theta = \frac{\pi}{8} ln2$$.

I could do this if instead of

$$ \int_0^{\frac{\pi}{4}} \ln(1+ \tan \theta)\ d\theta =\int_0^{\frac{\pi}{4}} \ln(1+ \tan \phi) - \ln2\ d\phi$$

I had

$$ \int_0^{\frac{\pi}{4}} \ln(1+ \tan \theta)\ d\theta = \int_0^{\frac{\pi}{4}} \ln2 - \ln(1+ \tan \phi)\ d\phi$$

Because both

$$\int_0^{\frac{\pi}{4}} \ln(1+ \tan \theta)\ d\theta $$ and $$\int_0^{\frac{\pi}{4}} \ln(1+ \tan \phi)\ d\phi$$

bound the same area and therefore have the same numerical value.

So I could say that

$$\int_0^{\frac{\pi}{4}} \ln(1+ \tan \theta)\ d\theta = \int_0^{\frac{\pi}{4}} \ln(1+ \tan \phi)\ d\phi$$

$$ \Rightarrow \int_0^{\frac{\pi}{4}} \ln(1+ \tan \phi)\ d\phi = \int_0^{\frac{\pi}{4}} \ln2 - \ln(1+ \tan \phi)\ d\phi$$

$$ \Rightarrow \int_0^{\frac{\pi}{4}} \ln(1+ \tan \phi)\ d\phi = \int_0^{\frac{\pi}{4}} \ln2 - \int_0^{\frac{\pi}{4}} \ln(1+ \tan \phi)\ d\phi$$

$$ \Rightarrow 2\int_0^{\frac{\pi}{4}} \ln(1+ \tan \phi)\ d\phi = \int_0^{\frac{\pi}{4}} \ln2\ d\phi$$

$$ \Rightarrow \int_0^{\frac{\pi}{4}} \ln(1+ \tan \phi)\ d\phi = \frac{1}{2} \int_0^{\frac{\pi}{4}} \ln2\ d\phi$$

$$ \Rightarrow \int_0^{\frac{\pi}{4}} \ln(1+ \tan \phi)\ d\phi = \frac{1}{2} [\phi \ln2]_0^{\frac{\pi}{4}}$$

$$ \Rightarrow \int_0^{\frac{\pi}{4}} \ln(1+ \tan \phi)\ d\phi = \frac{1}{2} [\frac{\pi}{4} \ln2 - 0\ln2]$$

$$ \Rightarrow \int_0^{\frac{\pi}{4}} \ln(1+ \tan \phi)\ d\phi = \frac{1}{2} [\frac{\pi}{4} \ln2] $$

$$ \Rightarrow \int_0^{\frac{\pi}{4}} \ln(1+ \tan \phi)\ d\phi = \frac{\pi}{8} \ln2 $$

Since that would all work so nicely, I am convinced that I must have missed out a minus sign at some point but I'm not sure where. Was it when I switched the limits between which I had to integrate?

If my mistake is missing a minus please tell me; if it's some other mistake then could you please tell me that it's not the minus problem but don't tell me what I have actually done wrong. I'd like to try and do as much of this by myself as possible. Thanks 92.1.69.95 (talk) 12:37, 7 August 2008 (UTC)


 * Near the beginning of your working, just after your substitution, you have


 * $$-\int_{\frac{\pi}{4}}^0 \ln(1+ \tan (\frac{\pi}{4} -\phi))\ d\phi

=-\int_0^{\frac{\pi}{4}} \ln(1+ \tan (\frac{\pi}{4} -\phi))\ d\phi$$


 * If you swap limits you must multiply the integral by -1 as well. so this should be


 * $$-\int_{\frac{\pi}{4}}^0 \ln(1+ \tan (\frac{\pi}{4} -\phi))\ d\phi

=\int_0^{\frac{\pi}{4}} \ln(1+ \tan (\frac{\pi}{4} -\phi))\ d\phi$$


 * which gives you the reversal in signs that you are looking for. Gandalf61 (talk) 13:13, 7 August 2008 (UTC)

Exponentially losing viewers
A newspaper story yesterday said a TV show "has, over the past few seasons, been exponentially losing viewers".

The story is HERE, but it is not necessary to read the story in order to understand the question.

I know that the word "exponentially" has come into wide use to mean "very rapid increase", whether or not the increase is exponential in the mathematical sense of the word.

What I'm wondering is whether the idea of "exponential decline" makes any sense mathematically. Can a finite number (such as number of viewers) which can at most decrease to zero, decrease exponentially?

Thanks, Wanderer57 (talk) 13:44, 7 August 2008 (UTC)


 * It can certainly do so approximately, as radioactive particles do (see exponential decay). It would mean that the rate of loss of viewers was constantly decreasing, in such a way that the half-life remains constant. Of course, this would break down with the last few people, but it could be a sensible approximation until then. I expect that article's just talking nonsense though. Algebraist 13:54, 7 August 2008 (UTC)


 * Interestingly, the exponential decay of particles also breaks down for the last few particles.  The chances that the writer actually ran the data through a curve-fitting program and determined that the resulting curve approximates an exponential rate of decay is almost zero.  However, I suppose we could come up with a reason for a non-linear rate of decline, if we suppose that everyone who stops watching tells their freinds it's now uncool to watch, and those friends then stop watching and tell their friends.  I would guess that a decreasing S-curve would be more likely to match the rate of decline, as that would start with a plateau, then have a slight rate of decrease, then a large rate of decrease, then again a slow rate of decrease, dropping to a lower plateau of loyal viewers who will stay with the program no matter what.  That lower level may or may not be enough viewers for the program to remain profitable. StuRat (talk) 14:12, 7 August 2008 (UTC)


 * Thank you both for the feedback. If I understand this correctly, a show that was "losing viewers exponentially" could be losing viewers at a very high rate (say 50% per week) OR losing viewers very gradually (1% per year), OR anything in between. To put this another way, "losing viewers exponentially" by itself says very little about the rate of loss.


 * Did I get that right? Wanderer57 (talk) 17:13, 7 August 2008 (UTC)
 * Yes. All it says is that the rate of loss varies proportionally to the number of viewers left. Algebraist 17:15, 7 August 2008 (UTC)


 * It is worth noting that "exponentially" is often used incorrectly by the lay public as a synonym for "fast". Hence when a news reporter says something like "exponentially losing viewers" I tend to assume what they really mean is "rapidly losing viewers".  Dragons flight (talk) 18:25, 7 August 2008 (UTC)
 * The decrease was monitored only the last few seasons, and it is not certain that the rate of decrease will remain contant in the future. It may not be a good idea to write that we lost a certain number of viewers per year, implying a linear decay, because a linearly decreasing function must from a certain point of time become negative, while the number of viewers cannot be a negative number. It may be better to write that we lost a certain fraction or percentage of the viewers each year, implying an exponential decay, although an exponentially decreasing function take fractional values and never becomes zero, while the number of viewers take integer values only and might eventually become zero. But 'linear' or 'exponential' says nothing about the actual decay, only about how to describe it, and about how to forecast the future number of viewers based on it. Bo Jacoby (talk) 22:20, 7 August 2008 (UTC).

Matrices
Can anyone tell me the history of matrices: who invented them and who found out their magical properties regarding application to certain physical problems? —Preceding unsigned comment added by 79.76.225.183 (talk) 23:24, 7 August 2008 (UTC)
 * Since you linked to the matrix disambiguation page, I will assume you haven't looked at the article: matrix (mathematics). Come back if you still have questions after reading the article.  J kasd  00:55, 8 August 2008 (UTC)


 * Im interested in the philosophy of matrices and why they appear to be so powerful in soling all sorts of problems. Is it just coincidence that this construct is so universally useful? —Preceding unsigned comment added by 79.76.225.183 (talk) 01:30, 8 August 2008 (UTC)
 * Matrices are really just a way of representing linear transformations between vector spaces. That's the reason, for example, for the (on its face, rather peculiar!) rule for multiplying them—it corresponds to taking the composition of the corresponding linear transformations (that is, do one transformation, then the other). --Trovatore (talk) 01:33, 8 August 2008 (UTC)


 * ... and the reason that matrices (and their cousins tensors) are frequently used to solve problems in physics is that the first thing a physicist does when confronted with a problem is to try to linearise it. And a linear mathematical model can be represented by vectors, matrices and tensors. The linear model may not be a complete solution - it may be a local solution which depends on assumptions that are only valid for a certain range of values of the problem's parameters. But the physicist hopes that by patching together linear solutions they may be able to at least develop solutions for a variety of practical cases. Non-linear problems exhibit unpredictable behaviours and tend to be much more difficult to solve. Gandalf61 (talk) 10:15, 8 August 2008 (UTC)


 * Lovely answer. Thanks —Preceding unsigned comment added by 79.76.208.36 (talk) 21:04, 8 August 2008 (UTC)